Chapter 5, Problem 22P

(a)

To determine

The slit width needed to resolve the interference pattern of the electrons if their kinetic energy is $0.010\text{ MeV}$ .

Answer to Problem 22P

The slit width needed to resolve the interference pattern of the electrons if their kinetic energy becomes $0.010\text{ MeV}$ is $0.0961\text{ nm}$ .

Explanation of Solution

Write the relativistic expression for the kinetic energy.

  $K=\left(\gamma -1\right)m{c}^2$

Here, $K$ is the kinetic energy, $\gamma$ is the Lorentz factor, $m$ is the mass of the particle and $c$ is the speed of light in vacuum.

Rewrite the above equation for $\gamma$ .

  $\begin{array}{c}\gamma -1=\frac{K}{m{c}^2}\\ \gamma =\frac{K}{m{c}^2}+1\end{array}$        (I)

Write the equation for the Lorentz factor.

  $\gamma =\frac{1}{1-v^2/c^2}$

Here, $v$ is the speed of the electron.

Rewrite the above equation for $v$ .

  $v=c{\left(1-\frac{1}{{\gamma }^2}\right)}^{1/2}$        (II)

Write the equation for the momentum.

  $p=\gamma mv$        (III)

Here, $p$ is the momentum of the electron.

Write the uncertainty principle.

  $\Delta x\Delta p\ge \frac{\hslash }{2}$        (IV)

Here, $\Delta x$ is the uncertainty in position, $\Delta p$ is the uncertainty in momentum and $\hslash$ is the reduced Planck’s constant.

The uncertainty in position must be equal to the slit width and since it is required to resolve a one percent difference in momentum, the value of $\Delta p$ must be $0.01p$ .

Substitute $a$ for $\Delta x$ and $0.01p$ for $\Delta p$ in equation (IV).

  $\begin{array}{c}\left(a\right)\left(0.01p\right)\ge \frac{\hslash }{2}\\ 0.01pa=\frac{\hslash }{2}\\ a=\frac{\hslash }{2\left(0.01p\right)}\end{array}$        (V)

Conclusion:

The mass of the electron is $9.11\times {10}^{-31}\text{ kg}$ , the value of $c$ is $3.0\times {10}^8\text{ m/s}$ , the value of $m{c}^2$ for electron is $0.511\text{ MeV}$ and the value of $\hslash$ is $1.054\times {10}^{-34}\text{ J}\cdot \text{s}$ .

Substitute $0.01\text{ MeV}$ for $K$ and $0.511\text{ MeV}$ for $m{c}^2$ in equation (I) to find $\gamma$ .

  $\begin{array}{c}\gamma =\frac{0.01\text{ MeV}}{0.511\text{ MeV}}+1\\ =1.02\end{array}$

Substitute $3.0\times {10}^8\text{ m/s}$ for $c$ and $1.02$ for $\gamma$ in equation (II) to find $v$ .

  $\begin{array}{c}v=\left(3.0\times {10}^8\text{ m/s}\right){\left(1-\frac{1}{{\left(1.02\right)}^2}\right)}^{1/2}\\ =5.91\times {10}^7\text{ m/s}\end{array}$

Substitute $1.02$ for $\gamma$ , $9.11\times {10}^{-31}\text{ kg}$ for $m$ and $5.91\times {10}^7\text{ m/s}$ for $v$ in equation (III) to find $p$ .

  $\begin{array}{c}p=\left(1.02\right)\left(9.11\times {10}^{-31}\text{ kg}\right)\left(5.91\times {10}^7\text{ m/s}\right)\\ =5.49\times {10}^{-23}\text{ kg}\cdot \text{m/s}\end{array}$

Substitute $1.054\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\hslash$ and $5.49\times {10}^{-23}\text{ kg}\cdot \text{m/s}$ for $p$ in equation (V) to find $a$ .

  $\begin{array}{c}a=\frac{1.054\times {10}^{-34}\text{ J}\cdot \text{s}}{2\left(0.01\right)\left(5.49\times {10}^{-23}\text{ kg}\cdot \text{m/s}\right)}\\ =0.0961\times {10}^{-9}\text{ m}\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\\ =0.0961\text{ nm}\end{array}$

Therefore, the slit width needed to resolve the interference pattern of the electrons if their kinetic energy becomes $0.010\text{ MeV}$ is $0.0961\text{ nm}$ .

(b)

To determine

The slit width needed to resolve the interference pattern of the electrons if their kinetic energy is $1.0\text{ MeV}$ .

Answer to Problem 22P

The slit width needed to resolve the interference pattern of the electrons if their kinetic energy becomes $1.0\text{ MeV}$ is $0.00694\text{ nm}$ .

Explanation of Solution

The value of the slit width can be determined in the same way used in part (a).

Conclusion:

Substitute $1.0\text{ MeV}$ for $K$ and $0.511\text{ MeV}$ for $m{c}^2$ in equation (I) to find $\gamma$ .

  $\begin{array}{c}\gamma =\frac{1.0\text{ MeV}}{0.511\text{ MeV}}+1\\ =2.96\end{array}$

Substitute $3.0\times {10}^8\text{ m/s}$ for $c$ and $2.96$ for $\gamma$ in equation (II) to find $v$ .

  $\begin{array}{c}v=\left(3.0\times {10}^8\text{ m/s}\right){\left(1-\frac{1}{{\left(2.96\right)}^2}\right)}^{1/2}\\ =2.82\times {10}^8\text{ m/s}\end{array}$

Substitute $2.96$ for $\gamma$ , $9.11\times {10}^{-31}\text{ kg}$ for $m$ and $2.82\times {10}^8\text{ m/s}$ for $v$ in equation (III) to find $p$ .

  $\begin{array}{c}p=\left(2.96\right)\left(9.11\times {10}^{-31}\text{ kg}\right)\left(2.82\times {10}^8\text{ m/s}\right)\\ =7.60\times {10}^{-22}\text{ kg}\cdot \text{m/s}\end{array}$

Substitute $1.054\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\hslash$ and $7.60\times {10}^{-22}\text{ kg}\cdot \text{m/s}$ for $p$ in equation (V) to find $a$ .

  $\begin{array}{c}a=\frac{1.054\times {10}^{-34}\text{ J}\cdot \text{s}}{2\left(0.01\right)\left(7.60\times {10}^{-22}\text{ kg}\cdot \text{m/s}\right)}\\ =0.00694\times {10}^{-9}\text{ m}\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\\ =0.00694\text{ nm}\end{array}$

Therefore, the slit width needed to resolve the interference pattern of the electrons if their kinetic energy becomes $1.0\text{ MeV}$ is $0.00694\text{ nm}$ .

(c)

To determine

The slit width needed to resolve the interference pattern of the electrons if their kinetic energy is $100\text{ MeV}$ .

Answer to Problem 22P

The slit width needed to resolve the interference pattern of the electrons if their kinetic energy becomes $100\text{ MeV}$ is $9.80\times {10}^{-14}\text{ nm}$ .

Explanation of Solution

The value of the slit width can be determined in the same way used in part (a).

Conclusion:

Substitute $100\text{ MeV}$ for $K$ and $0.511\text{ MeV}$ for $m{c}^2$ in equation (I) to find $\gamma$ .

  $\begin{array}{c}\gamma =\frac{100\text{ MeV}}{0.511\text{ MeV}}+1\\ =197\end{array}$

Substitute $3.0\times {10}^8\text{ m/s}$ for $c$ and $197$ for $\gamma$ in equation (II) to find $v$ .

  $\begin{array}{c}v=\left(3.0\times {10}^8\text{ m/s}\right){\left(1-\frac{1}{{\left(197\right)}^2}\right)}^{1/2}\\ =3.00\times {10}^8\text{ m/s}\end{array}$

Substitute $197$ for $\gamma$ , $9.11\times {10}^{-31}\text{ kg}$ for $m$ and $3.00\times {10}^8\text{ m/s}$ for $v$ in equation (III) to find $p$ .

  $\begin{array}{c}p=\left(197\right)\left(9.11\times {10}^{-31}\text{ kg}\right)\left(3.00\times {10}^8\text{ m/s}\right)\\ =5.38\times {10}^{-20}\text{ kg}\cdot \text{m/s}\end{array}$

Substitute $1.054\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\hslash$ and $5.38\times {10}^{-20}\text{ kg}\cdot \text{m/s}$ for $p$ in equation (V) to find $a$ .

  $\begin{array}{c}a=\frac{1.054\times {10}^{-34}\text{ J}\cdot \text{s}}{2\left(0.01\right)\left(5.38\times {10}^{-20}\text{ kg}\cdot \text{m/s}\right)}\\ =9.80\times {10}^{-23}\text{ m}\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\\ =9.80\times {10}^{-14}\text{ nm}\end{array}$

Therefore, the slit width needed to resolve the interference pattern of the electrons if their kinetic energy becomes $100\text{ MeV}$ is $9.80\times {10}^{-14}\text{ nm}$ .